Gautam Sethi <gautamsethi@gmail.com> wrote in message <cd7ccb2d-8c83-4568-a565-22f22a9416bf@googlegroups.com>...> Dear all:> I'm trying to find the end-points for runs of numbers based on a vector. Let> > >> Y = [1 3 4 6 7 8]> > Y => > 1 3 4 6 7 8> > I want an output vector Z of size(length(Y),2) where > > Z => > 0 2> 2 5> 2 5> 5 9> 5 9> 5 9> > Here is the idea behind what I want.> > The first element of Y is 1, which is not followed by the next integer; thus, the integers that "surround" the integer 1 are 0 and 2, the first row of Z.> > The second element of Y is 3, which is followed by the next integer, 4. However there is a break after 4. Thus, the run of consecutive integers 3 and 4 is "surrounded" by the integers 2 and 5. Therefore, the next two rows of Z, corresponding to the two elements 3 and 4 of Y, are [2 5].> > Likewise, the last three elements of Y are consecutive integers, "surrounded" by the integers 5 and 9. Hence the last three rows of Z are [5 9].> > Thanks for your help!- - - - - - - - - Y = Y'; f = find([true;diff(Y)>1;true]); Z = zeros(size(Y,1),2); Z(1,:) = [Y(1)-1,Y(f(2)-1)+1]; Z(f(2:end-1),:) = diff([Y(f(1:end-1)),Y(f(2:end)-1)],1,1); Z = cumsum(Z,1);